1490B - Balanced Remainders.cpp

/*
You are given a number n (divisible by 3) and an array a[1…n]. In one move, you can increase any of the array elements by one. Formally, you choose the index i (1=i=n) and replace ai with ai+1. You can choose the same index i multiple times for different moves.

Let's denote by c0, c1 and c2 the number of numbers from the array a that have remainders 0, 1 and 2 when divided by the number 3, respectively. Let's say that the array a has balanced remainders if c0, c1 and c2 are equal.

For example, if n=6 and a=[0,2,5,5,4,8], then the following sequence of moves is possible:

initially c0=1, c1=1 and c2=4, these values are not equal to each other. Let's increase a3, now the array a=[0,2,6,5,4,8];
c0=2, c1=1 and c2=3, these values are not equal. Let's increase a6, now the array a=[0,2,6,5,4,9];
c0=3, c1=1 and c2=2, these values are not equal. Let's increase a1, now the array a=[1,2,6,5,4,9];
c0=2, c1=2 and c2=2, these values are equal to each other, which means that the array a has balanced remainders.
Find the minimum number of moves needed to make the array a have balanced remainders.

Input
The first line contains one integer t (1=t=104). Then t test cases follow.

The first line of each test case contains one integer n (3=n=3·104) — the length of the array a. It is guaranteed that the number n is divisible by 3.

The next line contains n integers a1,a2,…,an (0=ai=100).

It is guaranteed that the sum of n over all test cases does not exceed 150000.

Output
For each test case, output one integer — the minimum number of moves that must be made for the a array to make it have balanced remainders.

Example
inputCopy
4
6
0 2 5 5 4 8
6
2 0 2 1 0 0
9
7 1 3 4 2 10 3 9 6
6
0 1 2 3 4 5
outputCopy
3
1
3
0
Note
The first test case is explained in the statements.

In the second test case, you need to make one move for i=2.

The third test case you need to make three moves:

the first move: i=9;
the second move: i=9;
the third move: i=2.
In the fourth test case, the values c0, c1 and c2 initially equal to each other, so the array a already has balanced remainders.
*/



#include<bits/stdc++.h>
using namespace std;

int main(){
    ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int t;
	cin >> t;
	
	while (t--) {
		int n;
		cin >> n;
		
		vector <int> rem(3, 0);
		
		for (int i = 0; i < n; i++) {
			int a;
			cin >> a;
			rem[a % 3]++;
		}
		
		int res = 0;
		
		while (rem[0] != rem[1] || rem[1] != rem[2]) {
			for (int i = 0; i < 3; i++) {
				if (rem[i] > n / 3) {
					rem[i]--;
					rem[(i + 1) % 3]++;
					res++;
				}
			}
		}
		
		cout << res << "\n";
	}

    return 0;
}